a = amount invested at 7%
b = amount invested at 9%
we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000. We can also say that

since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.
![\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20a%20%2B%20b%20%3D%2036000%5C%5C%5C%5C%200.07a%2B0.09b%3D2920%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20the%201st%20equation%7D%7D%7Ba%20%2B%20b%20%3D%2036000%5Cimplies%20%5Cunderline%7Bb%20%3D%2036000-a%7D%7D~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20on%20the%202nd%20equation%7D%7D%7B0.07a~~%20%2B%20~~0.09%28%5Cunderline%7B36000-a%7D%29~~%20%3D%20~~2920%7D%20%5C%5C%5C%5C%5C%5C%200.07a%2B3240-0.09a%3D2920%5Cimplies%203240-0.02a%3D2920%5Cimplies%20-0.02a%3D-320%20%5C%5C%5C%5C%5C%5C%20a%3D%5Ccfrac%7B-320%7D%7B-0.02%7D%5Cimplies%20%5Cboxed%7Ba%3D16000%7D~%5Chfill%20%5Cboxed%7B%5Cstackrel%7B36000~~%20-%20~~16000%7D%7B20000%3Db%7D%7D)
The answer would be D
-1 3/7=-10/7
-3 2/3=-11/3
-10/7 x -11/3=110/21
110/21=5 5/21
A negative times a negative =a positive
Answer:
There is nothing there...
Step-by-step explanation:
I wish I could help you but you need to put in a picture or diagram
The formula for volume always involves three dimensions.
In this case, the volume of a cube is
, where
is the side length of the cube.
To solve for that side length, you'd need to use a cube root.
Answer:
The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence level of
, we have the following confidence interval of proportions.

In which
z is the z-score that has a p-value of
.
A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.
This means that 
95% confidence level
So
, z is the value of Z that has a p-value of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).